What is the extreme f(x)= 1/(sqrt(25-x^2)) ?

The extreme f(x)= 1/(sqrt(25-x^2)) is Minimum(0, 1/5)

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extreme f(x)= 1/(sqrt(25-x^2))

extreme points

f (x) = \frac{1}{25 - x ^{2}}

Minimum (0, \frac{1}{5})

Solution steps

f^{'} (x) = \frac{x}{( 25 - x ^{2} ) ^{\frac{3}{2}}}

Find intervals:

Decreasing

: - 5 < x < 0,

Increasing

: 0 < x < 5

Plug

x = 0

into

\frac{1}{25 - x ^{2}} : \frac{1}{5}

Minimum (0, \frac{1}{5})

f (x) = 4 x^{3} - 3 x^{2} - 72 x - 80

x^{2} - 6 x + 3 y^{3} - y + 8

f (x, y) = 3 x^{2} + 4 x y + y^{2} + x^{3}

- sin (2 x)

f (x) = - 2 x^{3} + 42 x^{2} - 270 x + 3

What is the extreme f(x)= 1/(sqrt(25-x^2)) ?
The extreme f(x)= 1/(sqrt(25-x^2)) is Minimum(0, 1/5)